06 circular motion and other applications of newtons laws tủ tài liệu bách khoa

P U Z Z L E R
This sky diver is falling at more than
50 m/s (120 mi/h), but once her parachute opens, her downward velocity will
be greatly reduced. Why does she slow
down rapidly when her chute opens, enabling her to fall safely to the ground? If
the chute does not function properly, the
sky diver will almost certainly be seriously injured. What force exerted on
her limits her maximum speed?
(Guy Savage/Photo Researchers, Inc.)

c h a p t e r

Circular Motion and Other
Applications of Newton’s Laws
Chapter Outline
6.1 Newton’s Second Law Applied to
Uniform Circular Motion

6.2 Nonuniform Circular Motion
6.3 (Optional) Motion in Accelerated

6.4 (Optional) Motion in the Presence
of Resistive Forces

6.5 (Optional) Numerical Modeling in
Particle Dynamics

Frames

151

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Circular Motion and Other Applications of Newton’s Laws

I

n the preceding chapter we introduced Newton’s laws of motion and applied
them to situations involving linear motion. Now we discuss motion that is
slightly more complicated. For example, we shall apply Newton’s laws to objects
traveling in circular paths. Also, we shall discuss motion observed from an accelerating frame of reference and motion in a viscous medium. For the most part, this
chapter is a series of examples selected to illustrate the application of Newton’s
laws to a wide variety of circumstances.

6.1

NEWTON’S SECOND LAW APPLIED TO
UNIFORM CIRCULAR MOTION

In Section 4.4 we found that a particle moving with uniform speed v in a circular
path of radius r experiences an acceleration ar that has a magnitude
ar ϭ

4.7

v2
r

The acceleration is called the centripetal acceleration because ar is directed toward
the center of the circle. Furthermore, ar is always perpendicular to v. (If there
were a component of acceleration parallel to v, the particle’s speed would be
changing.)
Consider a ball of mass m that is tied to a string of length r and is being
whirled at constant speed in a horizontal circular path, as illustrated in Figure 6.1.
Its weight is supported by a low-friction table. Why does the ball move in a circle?
Because of its inertia, the tendency of the ball is to move in a straight line; however, the string prevents motion along a straight line by exerting on the ball a
force that makes it follow the circular path. This force is directed along the string
toward the center of the circle, as shown in Figure 6.1. This force can be any one
of our familiar forces causing an object to follow a circular path.
If we apply Newton’s second law along the radial direction, we find that the
value of the net force causing the centripetal acceleration can be evaluated:

⌺ Fr ϭ mar ϭ m

Force causing centripetal
acceleration

v2
r

(6.1)

m
Fr
r

Fr

Figure 6.1 Overhead view of a ball moving
in a circular path in a horizontal plane. A
force Fr directed toward the center of the circle keeps the ball moving in its circular path.


6.1

Newton’s Second Law Applied to Uniform Circular Motion

153

Figure 6.2 When the string breaks, the
ball moves in the direction tangent to the
circle.
r

A force causing a centripetal acceleration acts toward the center of the circular
path and causes a change in the direction of the velocity vector. If that force
should vanish, the object would no longer move in its circular path; instead, it
would move along a straight-line path tangent to the circle. This idea is illustrated
in Figure 6.2 for the ball whirling at the end of a string. If the string breaks at
some instant, the ball moves along the straight-line path tangent to the circle at
the point where the string broke.

Quick Quiz 6.1
Is it possible for a car to move in a circular path in such a way that it has a tangential acceleration but no centripetal acceleration?

CONCEPTUAL EXAMPLE 6.1

An athlete in the process of throwing the hammer at the 1996

Olympic Games in Atlanta, Georgia. The force exerted by the chain
is the force causing the circular
motion. Only when the athlete releases the hammer will it move
along a straight-line path tangent to
the circle.

Forces That Cause Centripetal Acceleration

The force causing centripetal acceleration is sometimes
called a centripetal force. We are familiar with a variety of forces
in nature — friction, gravity, normal forces, tension, and so
forth. Should we add centripetal force to this list?

Solution No; centripetal force should not be added to this
list. This is a pitfall for many students. Giving the force causing circular motion a name — centripetal force — leads many
students to consider it a new kind of force rather than a new
role for force. A common mistake in force diagrams is to draw
all the usual forces and then to add another vector for the
centripetal force. But it is not a separate force — it is simply
one of our familiar forces acting in the role of a force that causes
a circular motion.

Consider some examples. For the motion of the Earth
around the Sun, the centripetal force is gravity. For an object
sitting on a rotating turntable, the centripetal force is friction.
For a rock whirled on the end of a string, the centripetal
force is the force of tension in the string. For an amusementpark patron pressed against the inner wall of a rapidly rotating circular room, the centripetal force is the normal force exerted by the wall. What’s more, the centripetal force could
be a combination of two or more forces. For example, as a
Ferris-wheel rider passes through the lowest point, the centripetal force on her is the difference between the normal
force exerted by the seat and her weight.

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Circular Motion and Other Applications of Newton’s Laws

(a)

(b)

(c)

(d)

Figure 6.3 A ball that had been moving in a circular path is acted on by various external forces
that change its path.

Quick Quiz 6.2

QuickLab
Tie a string to a tennis ball, swing it in
a circle, and then, while it is swinging,
let go of the string to verify your answer to the last part of Quick Quiz 6.2.

A ball is following the dotted circular path shown in Figure 6.3 under the influence of a
force. At a certain instant of time, the force on the ball changes abruptly to a new force, and
the ball follows the paths indicated by the solid line with an arrowhead in each of the four
parts of the figure. For each part of the figure, describe the magnitude and direction of the

force required to make the ball move in the solid path. If the dotted line represents the
path of a ball being whirled on the end of a string, which path does the ball follow if
the string breaks?

Let us consider some examples of uniform circular motion. In each case, be
sure to recognize the external force (or forces) that causes the body to move in its
circular path.

EXAMPLE 6.2

How Fast Can It Spin?

A ball of mass 0.500 kg is attached to the end of a cord
1.50 m long. The ball is whirled in a horizontal circle as was
shown in Figure 6.1. If the cord can withstand a maximum
tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? Assume that the string remains
horizontal during the motion.

Solution It is difficult to know what might be a reasonable
value for the answer. Nonetheless, we know that it cannot be
too large, say 100 m/s, because a person cannot make a ball
move so quickly. It makes sense that the stronger the cord,
the faster the ball can twirl before the cord breaks. Also, we
expect a more massive ball to break the cord at a lower
speed. (Imagine whirling a bowling ball!)
Because the force causing the centripetal acceleration in
this case is the force T exerted by the cord on the ball, Equation 6.1 yields for ⌺Fr ϭ mar
v2
Tϭm
r

EXAMPLE 6.3

Solving for v, we have
vϭ

√

Tr
m

This shows that v increases with T and decreases with larger
m, as we expect to see — for a given v, a large mass requires a
large tension and a small mass needs only a small tension.
The maximum speed the ball can have corresponds to the
maximum tension. Hence, we find
vmax ϭ

√

Tmaxr
ϭ
m

√

(50.0 N)(1.50 m)
0.500 kg

ϭ 12.2 m/s

Exercise

Calculate the tension in the cord if the speed of
the ball is 5.00 m/s.

Answer

8.33 N.

The Conical Pendulum

A small object of mass m is suspended from a string of length
L. The object revolves with constant speed v in a horizontal
circle of radius r, as shown in Figure 6.4. (Because the string
sweeps out the surface of a cone, the system is known as a
conical pendulum.) Find an expression for v.

Let us choose ␪ to represent the angle between
string and vertical. In the free-body diagram shown in Figure
6.4, the force T exerted by the string is resolved into a vertical
component T cos ␪ and a horizontal component T sin ␪ acting toward the center of revolution. Because the object does

Solution


6.1

not accelerate in the vertical direction, ⌺ Fy ϭ may ϭ 0, and
the upward vertical component of T must balance the downward force of gravity. Therefore,

Because the force providing the centripetal acceleration in
this example is the component T sin ␪, we can use Newton’s
second law and Equation 6.1 to obtain

T cos ␪ ϭ mg

(1)

155

Newton’s Second Law Applied to Uniform Circular Motion

(2)

⌺ Fr ϭ T sin ␪ ϭ ma r ϭ

mv 2
r

Dividing (2) by (1) and remembering that sin ␪/cos ␪ ϭ
tan ␪, we eliminate T and find that
L θ

T cos θ

θ

T

r

From the geometry in Figure 6.4, we note that r ϭ L sin ␪ ;
therefore,
vϭ

mg
The conical pendulum and its free-body diagram.

EXAMPLE 6.4

√Lg sin ␪ tan ␪

Note that the speed is independent of the mass of the object.

What Is the Maximum Speed of the Car?

A 1 500-kg car moving on a flat, horizontal road negotiates a
curve, as illustrated in Figure 6.5. If the radius of the curve is
35.0 m and the coefficient of static friction between the tires

and dry pavement is 0.500, find the maximum speed the car
can have and still make the turn successfully.

Solution From experience, we should expect a maximum
speed less than 50 m/s. (A convenient mental conversion is
that 1 m/s is roughly 2 mi/h.) In this case, the force that enables the car to remain in its circular path is the force of static friction. (Because no slipping occurs at the point of contact between road and tires, the acting force is a force of
static friction directed toward the center of the curve. If this
force of static friction were zero — for example, if the car
were on an icy road — the car would continue in a straight

line and slide off the road.) Hence, from Equation 6.1 we
have

fs

(a)

n

fs

(a) The force of static friction directed toward the center of the curve keeps the car moving in a circular path. (b) The freebody diagram for the car.

fs ϭ m

(1)

v2
r

The maximum speed the car can have around the curve is
the speed at which it is on the verge of skidding outward. At
this point, the friction force has its maximum value
fs,max ϭ ␮sn. Because the car is on a horizontal road, the magnitude of the normal force equals the weight (n ϭ mg) and
thus fs,max ϭ ␮smg. Substituting this value for fs into (1), we
find that the maximum speed is

mg
(b)

Figure 6.5

v2
rg

v ϭ √rg tan ␪
T sin θ

mg

Figure 6.4

tan ␪ ϭ

vmax ϭ

√

fs,maxr
ϭ
m

√

␮smgr
ϭ √␮s gr
m

ϭ √(0.500)(9.80 m/s2)(35.0 m) ϭ 13.1 m/s

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Circular Motion and Other Applications of Newton’s Laws

Note that the maximum speed does not depend on the mass
of the car. That is why curved highways do not need multiple
speed limit signs to cover the various masses of vehicles using
the road.

EXAMPLE 6.5

Exercise

On a wet day, the car begins to skid on the curve
when its speed reaches 8.00 m/s. What is the coefficient of
static friction in this case?

Answer

0.187.

The Banked Exit Ramp

A civil engineer wishes to design a curved exit ramp for a
highway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a
car moving at the designated speed can negotiate the curve
even when the road is covered with ice. Such a ramp is usually banked; this means the roadway is tilted toward the inside

of the curve. Suppose the designated speed for the ramp is to
be 13.4 m/s (30.0 mi/h) and the radius of the curve is
50.0 m. At what angle should the curve be banked?

Solution On a level (unbanked) road, the force that
causes the centripetal acceleration is the force of static friction between car and road, as we saw in the previous example. However, if the road is banked at an angle ␪, as shown in
Figure 6.6, the normal force n has a horizontal component

n sin ␪ pointing toward the center of the curve. Because the
ramp is to be designed so that the force of static friction is
zero, only the component n sin ␪ causes the centripetal acceleration. Hence, Newton’s second law written for the radial direction gives

⌺ Fr ϭ n sin ␪ ϭ

(1)

The car is in equilibrium in the vertical direction. Thus, from
⌺Fy ϭ 0, we have
n cos ␪ ϭ mg

(2)

Dividing (1) by (2) gives
tan ␪ ϭ

v2
rg

␪ ϭ tanϪ1
n

θ
n cos θ

n sin θ

θ
mg

mg

Car rounding a curve on a road banked at an angle ␪
to the horizontal. When friction is neglected, the force that causes
the centripetal acceleration and keeps the car moving in its circular
path is the horizontal component of the normal force. Note that n is
the sum of the forces exerted by the road on the wheels.

Figure 6.6

EXAMPLE 6.6

mv2
r

m/s)
ϭ
΄ (50.0(13.4
m)(9.80 m/s ) ΅
2

2

20.1°

If a car rounds the curve at a speed less than 13.4 m/s,
friction is needed to keep it from sliding down the bank (to
the left in Fig. 6.6). A driver who attempts to negotiate the
curve at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig.
6.6). The banking angle is independent of the mass of the vehicle negotiating the curve.

Exercise Write Newton’s second law applied to the radial
direction when a frictional force fs is directed down the bank,
toward the center of the curve.
Answer

n sin ␪ ϩ fs cos ␪ ϭ

mv 2
r

Satellite Motion

This example treats a satellite moving in a circular orbit
around the Earth. To understand this situation, you must
know that the gravitational force between spherical objects
and small objects that can be modeled as particles having

masses m1 and m 2 and separated by a distance r is attractive
and has a magnitude
m1m2

Fg ϭ G
r2


6.1

where G ϭ 6.673 ϫ 10Ϫ11 Nи m2/kg2. This is Newton’s law of
gravitation, which we study in Chapter 14.
Consider a satellite of mass m moving in a circular orbit
around the Earth at a constant speed v and at an altitude h
above the Earth’s surface, as illustrated in Figure 6.7. Determine the speed of the satellite in terms of G, h, RE (the radius
of the Earth), and ME (the mass of the Earth).

Solution The only external force acting on the satellite is
the force of gravity, which acts toward the center of the Earth

and keeps the satellite in its circular orbit. Therefore,
Fr ϭ Fg ϭ G

r
h

MEm
r2

From Newton’s second law and Equation 6.1 we obtain
G

v2
MEm

ϭm
2
r
r

Solving for v and remembering that the distance r from the
center of the Earth to the satellite is r ϭ RE ϩ h, we obtain
vϭ

(1)

RE

157

Newton’s Second Law Applied to Uniform Circular Motion

√

GME
ϭ
r

√

GME
RE ϩ h

If the satellite were orbiting a different planet, its velocity
would increase with the mass of the planet and decrease as

the satellite’s distance from the center of the planet increased.

Fg

Exercise
v
m

Figure 6.7 A satellite of mass m moving around the Earth at a constant speed v in a circular orbit of radius r ϭ RE ϩ h. The force Fg
acting on the satellite that causes the centripetal acceleration is the
gravitational force exerted by the Earth on the satellite.

EXAMPLE 6.7

A satellite is in a circular orbit around the Earth at
an altitude of 1 000 km. The radius of the Earth is equal to
6.37 ϫ 106 m, and its mass is 5.98 ϫ 1024 kg. Find the speed
of the satellite, and then find the period, which is the time it
needs to make one complete revolution.

Answer

7.36 ϫ 103 m/s; 6.29 ϫ 103 s = 105 min.

Let’s Go Loop-the-Loop!

A pilot of mass m in a jet aircraft executes a loop-the-loop, as
shown in Figure 6.8a. In this maneuver, the aircraft moves in
a vertical circle of radius 2.70 km at a constant speed of
225 m/s. Determine the force exerted by the seat on the pilot

(a) at the bottom of the loop and (b) at the top of the loop.
Express your answers in terms of the weight of the pilot mg.

celeration has a magnitude n bot Ϫ mg, Newton’s second law
for the radial direction combined with Equation 6.1 gives

⌺ Fr ϭ nbot Ϫ mg ϭ m
nbot ϭ mg ϩ m

Solution

We expect the answer for (a) to be greater than
that for (b) because at the bottom of the loop the normal
and gravitational forces act in opposite directions, whereas at
the top of the loop these two forces act in the same direction.
It is the vector sum of these two forces that gives the force of
constant magnitude that keeps the pilot moving in a circular
path. To yield net force vectors with the same magnitude, the
normal force at the bottom (where the normal and gravitational forces are in opposite directions) must be greater than
that at the top (where the normal and gravitational forces are
in the same direction). (a) The free-body diagram for the pilot at the bottom of the loop is shown in Figure 6.8b. The
only forces acting on him are the downward force of gravity
Fg ϭ mg and the upward force n bot exerted by the seat. Because the net upward force that provides the centripetal ac-

v2
r

΂

v2

v2
ϭ mg 1 ϩ
r
rg

΃

Substituting the values given for the speed and radius gives

΄

nbot ϭ mg 1 ϩ

(225 m/s)2
(2.70 ϫ 103 m)(9.80 m/s2)

΅ϭ

2.91mg

Hence, the magnitude of the force n bot exerted by the seat
on the pilot is greater than the weight of the pilot by a factor
of 2.91. This means that the pilot experiences an apparent
weight that is greater than his true weight by a factor of 2.91.
(b) The free-body diagram for the pilot at the top of the
loop is shown in Figure 6.8c. As we noted earlier, both the
gravitational force exerted by the Earth and the force n top exerted by the seat on the pilot act downward, and so the net
downward force that provides the centripetal acceleration has

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CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws

Figure 6.8 (a) An aircraft executes a loop-the-loop maneuver as
it moves in a vertical circle at constant speed. (b) Free-body diagram for the pilot at the bottom
of the loop. In this position the
pilot experiences an apparent
weight greater than his true
weight. (c) Free-body diagram for
the pilot at the top of the loop.

n bot
Top

A

ntop
mg

mg

(b)

(c)

Bottom
(a)

a magnitude n top ϩ mg. Applying Newton’s second law yields

⌺ Fr ϭ ntop ϩ mg ϭ m

v2
r

In this case, the magnitude of the force exerted by the seat
on the pilot is less than his true weight by a factor of 0.913,
and the pilot feels lighter.

Exercise

Determine the magnitude of the radially directed
force exerted on the pilot by the seat when the aircraft is at
point A in Figure 6.8a, midway up the loop.

΂ vrg Ϫ 1΃

ntop ϭ m

v2
Ϫ mg ϭ mg
r

2

ntop ϭ mg

m/s)
Ϫ 1΅ ϭ
΄ (2.70 ϫ (225
10 m)(9.80 m/s )
2

3

2

0.913mg

Answer

nA ϭ 1.913mg directed to the right.

Quick Quiz 6.3
A bead slides freely along a curved wire at constant speed, as shown in the overhead view of
Figure 6.9. At each of the points Ꭽ, Ꭾ, and Ꭿ, draw the vector representing the force that
the wire exerts on the bead in order to cause it to follow the path of the wire at that point.

Ꭽ
Ꭾ

QuickLab
Hold a shoe by the end of its lace and
spin it in a vertical circle. Can you
feel the difference in the tension in
the lace when the shoe is at top of the
circle compared with when the shoe

is at the bottom?

Ꭿ

6.2

Figure 6.9

NONUNIFORM CIRCULAR MOTION

In Chapter 4 we found that if a particle moves with varying speed in a circular
path, there is, in addition to the centripetal (radial) component of acceleration, a
tangential component having magnitude dv/dt. Therefore, the force acting on the


6.2

Nonuniform Circular Motion

159

Some examples of forces acting during circular motion. (Left) As these speed skaters round a
curve, the force exerted by the ice on their skates provides the centripetal acceleration.
(Right) Passengers on a “corkscrew” roller coaster. What are the origins of the forces in this
example?

Figure 6.10 When the force acting on a particle moving in a circular path has a tangential component Ft , the
particle’s speed changes. The total force exerted on the
particle in this case is the vector sum of the radial force
and the tangential force. That is, F ϭ Fr ϩ Ft .

F
Fr

Ft

particle must also have a tangential and a radial component. Because the total acceleration is a ϭ ar ϩ at , the total force exerted on the particle is F ϭ Fr ϩ Ft , as
shown in Figure 6.10. The vector Fr is directed toward the center of the circle and is
responsible for the centripetal acceleration. The vector Ft tangent to the circle is responsible for the tangential acceleration, which represents a change in the speed of
the particle with time. The following example demonstrates this type of motion.

EXAMPLE 6.8

Keep Your Eye on the Ball

A small sphere of mass m is attached to the end of a cord of
length R and whirls in a vertical circle about a fixed point O,
as illustrated in Figure 6.11a. Determine the tension in the
cord at any instant when the speed of the sphere is v and the
cord makes an angle ␪ with the vertical.

Solution Unlike the situation in Example 6.7, the speed is
not uniform in this example because, at most points along the
path, a tangential component of acceleration arises from the
gravitational force exerted on the sphere. From the free-body
diagram in Figure 6.11b, we see that the only forces acting on


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CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws
vtop

mg

Ttop

R
O

O
T

mg cos θ

T bot

θ

v bot

mg sin θ

θ

mg
mg
(a)

(b)

Figure 6.11 (a) Forces acting on a sphere
of mass m connected to a cord of length R and
rotating in a vertical circle centered at O.
(b) Forces acting on the sphere at the top and
bottom of the circle. The tension is a maximum at the bottom and a minimum at the top.

the sphere are the gravitational force Fg ϭ m g exerted by the
Earth and the force T exerted by the cord. Now we resolve Fg
into a tangential component mg sin ␪ and a radial component
mg cos ␪. Applying Newton’s second law to the forces acting
on the sphere in the tangential direction yields

Special Cases At the top of the path, where ␪ ϭ 180°, we
have cos 180° ϭ Ϫ 1, and the tension equation becomes

⌺ Ft ϭ mg sin ␪ ϭ mat

This is the minimum value of T. Note that at this point at ϭ 0
and therefore the acceleration is purely radial and directed
downward.
At the bottom of the path, where ␪ ϭ 0, we see that, because cos 0 ϭ 1,
v 2bot
Tbot ϭ m
ϩg
R

at ϭ g sin ␪
This tangential component of the acceleration causes v to

change in time because at ϭ dv/dt.
Applying Newton’s second law to the forces acting on the
sphere in the radial direction and noting that both T and ar
are directed toward O, we obtain
mv2
⌺ Fr ϭ T Ϫ mg cos ␪ ϭ R
Tϭ m

΂

v2
ϩ g cos ␪
R

Ttop ϭ m

΂

v 2top
R

΂

΃

Ϫg

΃

This is the maximum value of T. At this point, at is again 0

and the acceleration is now purely radial and directed upward.

Exercise

At what position of the sphere would the cord
most likely break if the average speed were to increase?

΃

Answer

At the bottom, where T has its maximum value.

Optional Section

6.3

MOTION IN ACCELERATED FRAMES

When Newton’s laws of motion were introduced in Chapter 5, we emphasized that
they are valid only when observations are made in an inertial frame of reference.
In this section, we analyze how an observer in a noninertial frame of reference
(one that is accelerating) applies Newton’s second law.


6.3

4.8

161

Motion in Accelerated Frames

To understand the motion of a system that is noninertial because an object is
moving along a curved path, consider a car traveling along a highway at a high
speed and approaching a curved exit ramp, as shown in Figure 6.12a. As the car
takes the sharp left turn onto the ramp, a person sitting in the passenger seat
slides to the right and hits the door. At that point, the force exerted on her by the
door keeps her from being ejected from the car. What causes her to move toward
the door? A popular, but improper, explanation is that some mysterious force acting from left to right pushes her outward. (This is often called the “centrifugal”
force, but we shall not use this term because it often creates confusion.) The passenger invents this fictitious force to explain what is going on in her accelerated
frame of reference, as shown in Figure 6.12b. (The driver also experiences this effect but holds on to the steering wheel to keep from sliding to the right.)
The phenomenon is correctly explained as follows. Before the car enters the
ramp, the passenger is moving in a straight-line path. As the car enters the ramp
and travels a curved path, the passenger tends to move along the original straightline path. This is in accordance with Newton’s first law: The natural tendency of a
body is to continue moving in a straight line. However, if a sufficiently large force
(toward the center of curvature) acts on the passenger, as in Figure 6.12c, she will
move in a curved path along with the car. The origin of this force is the force of
friction between her and the car seat. If this frictional force is not large enough,
she will slide to the right as the car turns to the left under her. Eventually, she encounters the door, which provides a force large enough to enable her to follow the
same curved path as the car. She slides toward the door not because of some mysterious outward force but because the force of friction is not sufficiently great
to allow her to travel along the circular path followed by the car.
In general, if a particle moves with an acceleration a relative to an observer in
an inertial frame, that observer may use Newton’s second law and correctly claim
that ⌺F ϭ ma. If another observer in an accelerated frame tries to apply Newton’s
second law to the motion of the particle, the person must introduce fictitious
forces to make Newton’s second law work. These forces “invented” by the observer
in the accelerating frame appear to be real. However, we emphasize that these fictitious forces do not exist when the motion is observed in an inertial frame.
Fictitious forces are used only in an accelerating frame and do not represent “real”
forces acting on the particle. (By real forces, we mean the interaction of the particle with its environment.) If the fictitious forces are properly defined in the accelerating frame, the description of motion in this frame is equivalent to the description given by an inertial observer who considers only real forces. Usually, we

analyze motions using inertial reference frames, but there are cases in which it is
more convenient to use an accelerating frame.

Figure 6.12 (a) A car approaching a curved exit ramp. What causes a front-seat passenger to
move toward the right-hand door? (b) From the frame of reference of the passenger, a (fictitious) force pushes her toward the right door. (c) Relative to the reference frame of the Earth,
the car seat applies a leftward force to the passenger, causing her to change direction along with
the rest of the car.

QuickLab
Use a string, a small weight, and a
protractor to measure your acceleration as you start sprinting from a
standing position.

Fictitious forces

(a)

(b)

(c)


162

CHAPTER 6

EXAMPLE 6.9

Circular Motion and Other Applications of Newton’s Laws

Fictitious Forces in Linear Motion

A small sphere of mass m is hung by a cord from the ceiling
of a boxcar that is accelerating to the right, as shown in Figure 6.13. According to the inertial observer at rest (Fig.
6.13a), the forces on the sphere are the force T exerted by
the cord and the force of gravity. The inertial observer concludes that the acceleration of the sphere is the same as that
of the boxcar and that this acceleration is provided by the
horizontal component of T. Also, the vertical component of
T balances the force of gravity. Therefore, she writes Newton’s second law as ⌺F ϭ T ϩ m g ϭ ma, which in component form becomes
Inertial observer

Ά

(1)
(2)

⌺ Fx ϭ T sin ␪ ϭ ma
⌺ Fy ϭ T cos ␪ Ϫ mg ϭ 0

Because the deflection of the cord from the vertical serves as
a measure of acceleration, a simple pendulum can be used as an
accelerometer.
According to the noninertial observer riding in the car
(Fig. 6.13b), the cord still makes an angle ␪ with the vertical;
however, to her the sphere is at rest and so its acceleration is
zero. Therefore, she introduces a fictitious force to balance
the horizontal component of T and claims that the net force
on the sphere is zero! In this noninertial frame of reference,
Newton’s second law in component form yields
Noninertial observer

Thus, by solving (1) and (2) simultaneously for a, the inertial
observer can determine the magnitude of the car’s acceleration through the relationship
a ϭ g tan ␪

Ά⌺⌺

F xЈ ϭ T sin ␪ Ϫ Ffictitious ϭ 0
F yЈ ϭ T cos ␪ Ϫ mg ϭ 0

If we recognize that Ffictitious ϭ ma inertial ϭ ma, then these expressions are equivalent to (1) and (2); therefore, the noninertial observer obtains the same mathematical results as the inertial observer does. However, the physical interpretation of the
deflection of the cord differs in the two frames of reference.

a

Inertial
observer

T θ
mg

(a)
Noninertial
observer

Ffictitious

T θ
mg

(b)

Figure 6.13 A small sphere suspended from the ceiling of a boxcar accelerating to the right is deflected as shown. (a) An inertial observer at rest outside the car claims that the acceleration of the
sphere is provided by the horizontal component of T. (b) A noninertial observer riding in the car says
that the net force on the sphere is zero and that the deflection of the cord from the vertical is due to a
fictitious force Ffictitious that balances the horizontal component of T.


6.4

EXAMPLE 6.10

Motion in the Presence of Resistive Forces

Fictitious Force in a Rotating System
According to a noninertial observer attached to the
turntable, the block is at rest and its acceleration is zero.
Therefore, she must introduce a fictitious outward force of
magnitude mv 2/r to balance the inward force exerted by the
string. According to her, the net force on the block is zero,
and she writes Newton’s second law as T Ϫ mv 2/r ϭ 0.

Suppose a block of mass m lying on a horizontal, frictionless
turntable is connected to a string attached to the center of
the turntable, as shown in Figure 6.14. According to an inertial observer, if the block rotates uniformly, it undergoes an
acceleration of magnitude v 2/r, where v is its linear speed.
The inertial observer concludes that this centripetal acceleration is provided by the force T exerted by the string and
writes Newton’s second law as T ϭ mv 2/r.

n

Noninertial observer

n

T

Ffictitious =

mg

T

mv 2
r

mg

(a)

Inertial observer

(b)

Figure 6.14 A block of mass m connected to a string tied to the center of a rotating turntable.
(a) The inertial observer claims that the force causing the circular motion is provided by the force T
exerted by the string on the block. (b) The noninertial observer claims that the block is not accelerating, and therefore she introduces a fictitious force of magnitude mv 2/r that acts outward and balances
the force T.

Optional Section

6.4
4.9

163

MOTION IN THE PRESENCE OF RESISTIVE FORCES

In the preceding chapter we described the force of kinetic friction exerted on an
object moving on some surface. We completely ignored any interaction between
the object and the medium through which it moves. Now let us consider the effect
of that medium, which can be either a liquid or a gas. The medium exerts a resistive force R on the object moving through it. Some examples are the air resistance associated with moving vehicles (sometimes called air drag) and the viscous
forces that act on objects moving through a liquid. The magnitude of R depends
on such factors as the speed of the object, and the direction of R is always opposite
the direction of motion of the object relative to the medium. The magnitude of R
nearly always increases with increasing speed.
The magnitude of the resistive force can depend on speed in a complex way,
and here we consider only two situations. In the first situation, we assume the resistive force is proportional to the speed of the moving object; this assumption is
valid for objects falling slowly through a liquid and for very small objects, such as
dust particles, moving through air. In the second situation, we assume a resistive
force that is proportional to the square of the speed of the moving object; large
objects, such as a skydiver moving through air in free fall, experience such a force.


164

CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws
v=0

a=g

v

R

vt

v

0.63vt

mg

(a)

t

τ

v = vt
a=0

(c)

(b)

Figure 6.15

(a) A small sphere falling through a liquid. (b) Motion diagram of the sphere as it

falls. (c) Speed – time graph for the sphere. The sphere reaches a maximum, or terminal, speed
vt , and the time constant ␶ is the time it takes to reach 0.63vt .

Resistive Force Proportional to Object Speed
If we assume that the resistive force acting on an object moving through a liquid
or gas is proportional to the object’s speed, then the magnitude of the resistive
force can be expressed as
R ϭ bv

(6.2)

where v is the speed of the object and b is a constant whose value depends on the
properties of the medium and on the shape and dimensions of the object. If the
object is a sphere of radius r, then b is proportional to r.
Consider a small sphere of mass m released from rest in a liquid, as in Figure
6.15a. Assuming that the only forces acting on the sphere are the resistive force bv
and the force of gravity Fg , let us describe its motion.1 Applying Newton’s second
law to the vertical motion, choosing the downward direction to be positive, and
noting that ⌺Fy ϭ mg Ϫ bv, we obtain
mg Ϫ bv ϭ ma ϭ m

dv
dt

(6.3)

where the acceleration dv/dt is downward. Solving this expression for the acceleration gives
b
dv
ϭgϪ

v
dt
m

Terminal speed

(6.4)

This equation is called a differential equation, and the methods of solving it may not
be familiar to you as yet. However, note that initially, when v ϭ 0, the resistive
force Ϫ bv is also zero and the acceleration dv/dt is simply g. As t increases, the resistive force increases and the acceleration decreases. Eventually, the acceleration
becomes zero when the magnitude of the resistive force equals the sphere’s
weight. At this point, the sphere reaches its terminal speed vt , and from then on
1

There is also a buoyant force acting on the submerged object. This force is constant, and its magnitude
is equal to the weight of the displaced liquid. This force changes the apparent weight of the sphere by a
constant factor, so we will ignore the force here. We discuss buoyant forces in Chapter 15.


6.4

165

Motion in the Presence of Resistive Forces

it continues to move at this speed with zero acceleration, as shown in Figure 6.15b.
We can obtain the terminal speed from Equation 6.3 by setting a ϭ dv/dt ϭ 0.
This gives
mg Ϫ bvt ϭ 0

or
vt ϭ mg/b
The expression for v that satisfies Equation 6.4 with v ϭ 0 at t ϭ 0 is
vϭ

mg
(1 Ϫ eϪbt/m) ϭ vt (1 Ϫ eϪt/␶)
b

(6.5)

This function is plotted in Figure 6.15c. The time constant ␶ ϭ m/b (Greek letter
tau) is the time it takes the sphere to reach 63.2% (ϭ 1 Ϫ 1/e) of its terminal
speed. This can be seen by noting that when t ϭ ␶, Equation 6.5 yields v ϭ 0.632vt .
We can check that Equation 6.5 is a solution to Equation 6.4 by direct differentiation:
mg d Ϫbt/m
mg Ϫbt/m
dv
d mg
ϭϪ
ϭ geϪbt/m
ϭ
Ϫ
e
e
dt
dt
b
b
b dt

΂

΃

(See Appendix Table B.4 for the derivative of e raised to some power.) Substituting
into Equation 6.4 both this expression for dv/dt and the expression for v given by
Equation 6.5 shows that our solution satisfies the differential equation.

EXAMPLE 6.11

Sphere Falling in Oil

A small sphere of mass 2.00 g is released from rest in a large
vessel filled with oil, where it experiences a resistive force proportional to its speed. The sphere reaches a terminal speed
of 5.00 cm/s. Determine the time constant ␶ and the time it
takes the sphere to reach 90% of its terminal speed.

0.900vt ϭ vt(1 Ϫ eϪt/␶ )
1 Ϫ eϪt/␶ ϭ 0.900
eϪt/␶ ϭ 0.100
Ϫ

Solution

Because the terminal
vt ϭ mg/b, the coefficient b is
bϭ

speed

is

given

by

Therefore, the time constant ␶ is
2.00 g
m
ϭ
ϭ 5.10 ϫ 10Ϫ3 s
b
392 g/s

The speed of the sphere as a function of time is given by
Equation 6.5. To find the time t it takes the sphere to reach a
speed of 0.900vt , we set v ϭ 0.900vt in Equation 6.5 and solve
for t:

t
ϭ ln(0.100) ϭ Ϫ2.30
␶
t ϭ 2.30␶ ϭ 2.30(5.10 ϫ 10Ϫ3 s) ϭ 11.7 ϫ 10Ϫ3 s

mg
(2.00 g)(980 cm/s2)
ϭ 392 g/s
ϭ
vt

5.00 cm/s

␶ϭ

Aerodynamic car. A streamlined
body reduces air drag and increases fuel efficiency.

ϭ 11.7 ms
Thus, the sphere reaches 90% of its terminal (maximum)
speed in a very short time.

Exercise What is the sphere’s speed through the oil at t ϭ
11.7 ms? Compare this value with the speed the sphere would
have if it were falling in a vacuum and so were influenced
only by gravity.
Answer

4.50 cm/s in oil versus 11.5 cm/s in free fall.

Air Drag at High Speeds
For objects moving at high speeds through air, such as airplanes, sky divers, cars,
and baseballs, the resistive force is approximately proportional to the square of the
speed. In these situations, the magnitude of the resistive force can be expressed as
R ϭ 12D␳Av2

(6.6)


166

CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws

where ␳ is the density of air, A is the cross-sectional area of the falling object measured in a plane perpendicular to its motion, and D is a dimensionless empirical
quantity called the drag coefficient. The drag coefficient has a value of about 0.5 for
spherical objects but can have a value as great as 2 for irregularly shaped objects.
Let us analyze the motion of an object in free fall subject to an upward air
resistive force of magnitude R ϭ 12 D␳Av2. Suppose an object of mass m is released from rest. As Figure 6.16 shows, the object experiences two external forces:
the downward force of gravity Fg ϭ mg and the upward resistive force R. (There is
also an upward buoyant force that we neglect.) Hence, the magnitude of the net
force is

R
v

R

⌺ F ϭ mg Ϫ 12 D␳Av2

mg
vt

mg

Figure 6.16 An object falling
through air experiences a resistive
force R and a gravitational force
Fg ϭ mg. The object reaches terminal speed (on the right) when the
net force acting on it is zero, that

is, when R ϭ Ϫ Fg or R ϭ mg. Before this occurs, the acceleration
varies with speed according to
Equation 6.8.

(6.7)

where we have taken downward to be the positive vertical direction. Substituting
⌺F ϭ ma into Equation 6.7, we find that the object has a downward acceleration of
magnitude
D␳A
aϭgϪ
v2
(6.8)
2m

΂

΃

We can calculate the terminal speed vt by using the fact that when the force of
gravity is balanced by the resistive force, the net force on the object is zero and
therefore its acceleration is zero. Setting a ϭ 0 in Equation 6.8 gives
gϪ

΂ D2m␳A ΃ v

t

2

ϭ0

vt ϭ

√

2mg
D␳A

(6.9)

Using this expression, we can determine how the terminal speed depends on the
dimensions of the object. Suppose the object is a sphere of radius r. In this case,
A ϰ r2 (from A ϭ ␲r 2 ) and m ϰ r3 (because the mass is proportional to the
volume of the sphere, which is V ϭ 43 ␲r3). Therefore, vt ϰ √r.
Table 6.1 lists the terminal speeds for several objects falling through air.

The high cost of fuel has prompted many truck owners to install wind deflectors on their cabs to
reduce drag.


6.4

Motion in the Presence of Resistive Forces

167

TABLE 6.1 Terminal Speed for Various Objects Falling Through Air
Object

Mass (kg)

Cross-Sectional Area
(m2)

vt (m/s)

Sky diver
Baseball (radius 3.7 cm)
Golf ball (radius 2.1 cm)
Hailstone (radius 0.50 cm)
Raindrop (radius 0.20 cm)

75
0.145
0.046
4.8 ϫ 10Ϫ4
3.4 ϫ 10Ϫ5

0.70
4.2 ϫ 10Ϫ3
1.4 ϫ 10Ϫ3
7.9 ϫ 10Ϫ5
1.3 ϫ 10Ϫ5

60
43
44
14
9.0

CONCEPTUAL EXAMPLE 6.12
Consider a sky surfer who jumps from a plane with her feet
attached firmly to her surfboard, does some tricks, and then
opens her parachute. Describe the forces acting on her during these maneuvers.

Solution When the surfer first steps out of the plane, she
has no vertical velocity. The downward force of gravity causes
her to accelerate toward the ground. As her downward speed
increases, so does the upward resistive force exerted by the
air on her body and the board. This upward force reduces
their acceleration, and so their speed increases more slowly.
Eventually, they are going so fast that the upward resistive
force matches the downward force of gravity. Now the net
force is zero and they no longer accelerate, but reach their
terminal speed. At some point after reaching terminal speed,
she opens her parachute, resulting in a drastic increase in the
upward resistive force. The net force (and thus the acceleration) is now upward, in the direction opposite the direction
of the velocity. This causes the downward velocity to decrease
rapidly; this means the resistive force on the chute also decreases. Eventually the upward resistive force and the downward force of gravity balance each other and a much smaller
terminal speed is reached, permitting a safe landing.
(Contrary to popular belief, the velocity vector of a sky
diver never points upward. You may have seen a videotape
in which a sky diver appeared to “rocket” upward once the
chute opened. In fact, what happened is that the diver
slowed down while the person holding the camera continued falling at high speed.)

EXAMPLE 6.13

A sky surfer takes advantage of the upward force of the air on her

board. (

Falling Coffee Filters

The dependence of resistive force on speed is an empirical
relationship. In other words, it is based on observation rather
than on a theoretical model. A series of stacked filters is
dropped, and the terminal speeds are measured. Table 6.2

presents data for these coffee filters as they fall through the
air. The time constant ␶ is small, so that a dropped filter
quickly reaches terminal speed. Each filter has a mass of
1.64 g. When the filters are nested together, they stack in


168

CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws

such a way that the front-facing surface area does not increase. Determine the relationship between the resistive force
exerted by the air and the speed of the falling filters.

Solution At terminal speed, the upward resistive force balances the downward force of gravity. So, a single filter falling
at its terminal speed experiences a resistive force of
1.64 g
(9.80 m/s ) ϭ 0.016 1 N
΂ 1000
g/kg ΃

R ϭ mg ϭ

2

Two filters nested together experience 0.032 2 N of resistive
force, and so forth. A graph of the resistive force on the filters as a function of terminal speed is shown in Figure 6.17a.
A straight line would not be a good fit, indicating that the resistive force is not proportional to the speed. The curved line
is for a second-order polynomial, indicating a proportionality
of the resistive force to the square of the speed. This proportionality is more clearly seen in Figure 6.17b, in which the resistive force is plotted as a function of the square of the terminal speed.

TABLE 6.2
Terminal Speed for
Stacked Coffee Filters
Number
of Filters

vt
(m/s)a

1
2
3
4
5
6
7
8
9
10

1.01
1.40
1.63
2.00
2.25
2.40
2.57
2.80
3.05
3.22

All values of vt are approximate.

0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00

Resistive force (N)

Resistive force (N)

a

Pleated coffee filters can be nested together so
that the force of air resistance can be studied.
(

0

1

2

3

4

0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.00

0

2

4

6

8

Terminal speed (m/s)

Terminal speed squared (m/s)2

(a)

(b)

Figure 6.17 (a) Relationship between the resistive force acting on falling coffee filters and their terminal speed. The curved line is a second-order polynomial fit. (b) Graph relating the resistive force to
the square of the terminal speed. The fit of the straight line to the data points indicates that the resistive force is proportional to the terminal speed squared. Can you find the proportionality constant?

10

12


6.5

EXAMPLE 6.14

Numerical Modeling in Particle Dynamics

169

Resistive Force Exerted on a Baseball

A pitcher hurls a 0.145-kg baseball past a batter at 40.2 m/s
(ϭ90 mi/h). Find the resistive force acting on the ball at this
speed.

Solution

We do not expect the air to exert a huge force
on the ball, and so the resistive force we calculate from Equation 6.6 should not be more than a few newtons. First, we
must determine the drag coefficient D. We do this by imagining that we drop the baseball and allow it to reach terminal
speed. We solve Equation 6.9 for D and substitute the appropriate values for m, vt , and A from Table 6.1. Taking the density of air as 1.29 kg/m3, we obtain

Dϭ

2 mg

vt2 ␳A
ϭ 0.284

ϭ

2(0.145 kg)(9.80 m/s2)
(43 m/s)2 (1.29 kg/m3)(4.2 ϫ 10Ϫ3 m2)

This number has no dimensions. We have kept an extra digit
beyond the two that are significant and will drop it at the end
of our calculation.
We can now use this value for D in Equation 6.6 to find
the magnitude of the resistive force:

R ϭ 12 D␳Av2
ϭ 12(0.284)(1.29 kg/m3)(4.2 ϫ 10Ϫ3 m2)(40.2 m/s)2
ϭ 1.2 N

Optional Section

6.5

NUMERICAL MODELING IN PARTICLE DYNAMICS 2

As we have seen in this and the preceding chapter, the study of the dynamics of a
particle focuses on describing the position, velocity, and acceleration as functions of
time. Cause-and-effect relationships exist among these quantities: Velocity causes
position to change, and acceleration causes velocity to change. Because acceleration is the direct result of applied forces, any analysis of the dynamics of a particle
usually begins with an evaluation of the net force being exerted on the particle.
Up till now, we have used what is called the analytical method to investigate the
position, velocity, and acceleration of a moving particle. Let us review this method
briefly before learning about a second way of approaching problems in dynamics.
(Because we confine our discussion to one-dimensional motion in this section,
boldface notation will not be used for vector quantities.)
If a particle of mass m moves under the influence of a net force ⌺F, Newton’s
second law tells us that the acceleration of the particle is a ϭ ⌺F/m. In general, we
apply the analytical method to a dynamics problem using the following procedure:
1.
2.
3.
4.

Sum all the forces acting on the particle to get the net force ⌺F.
Use this net force to determine the acceleration from the relationship a ϭ ⌺F/m.

Use this acceleration to determine the velocity from the relationship dv/dt ϭ a.
Use this velocity to determine the position from the relationship dx/dt ϭ v.
The following straightforward example illustrates this method.

EXAMPLE 6.15

An Object Falling in a Vacuum — Analytical Method

Consider a particle falling in a vacuum under the influence
of the force of gravity, as shown in Figure 6.18. Use the analytical method to find the acceleration, velocity, and position of
the particle.
2

Solution The only force acting on the particle is the
downward force of gravity of magnitude Fg , which is also the
net force. Applying Newton’s second law, we set the net force
acting on the particle equal to the mass of the particle times

The authors are most grateful to Colonel James Head of the U.S. Air Force Academy for preparing
this section. See the Student Tools CD-ROM for some assistance with numerical modeling.


170

CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws

its acceleration (taking upward to be the positive y direction):
Fg ϭ ma y ϭ Ϫmg

In these expressions, yi and vyi represent the position and
speed of the particle at t i ϭ 0.

Thus, a y ϭ Ϫg, which means the acceleration is constant. Because dv y /dt ϭ a y, we see that dv y /dt ϭ Ϫg, which may be integrated to yield
v y(t) ϭ v yi Ϫ gt
Then, because v y ϭ dy/dt, the position of the particle is obtained from another integration, which yields the well-known
result
y(t) ϭ y i ϩ v yi t Ϫ 12 gt 2

mg

Figure 6.18

An object falling in vacuum under the influence

of gravity.

The analytical method is straightforward for many physical situations. In the
“real world,” however, complications often arise that make analytical solutions difficult and perhaps beyond the mathematical abilities of most students taking introductory physics. For example, the net force acting on a particle may depend on
the particle’s position, as in cases where the gravitational acceleration varies with
height. Or the force may vary with velocity, as in cases of resistive forces caused by
motion through a liquid or gas.
Another complication arises because the expressions relating acceleration, velocity, position, and time are differential equations rather than algebraic ones. Differential equations are usually solved using integral calculus and other special
techniques that introductory students may not have mastered.
When such situations arise, scientists often use a procedure called numerical
modeling to study motion. The simplest numerical model is called the Euler
method, after the Swiss mathematician Leonhard Euler (1707 – 1783).

The Euler Method

In the Euler method for solving differential equations, derivatives are approximated as ratios of finite differences. Considering a small increment of time ⌬t, we
can approximate the relationship between a particle’s speed and the magnitude of
its acceleration as
a(t) Ϸ

⌬v
v(t ϩ ⌬t) Ϫ v(t)
ϭ
⌬t
⌬t

Then the speed v(t ϩ ⌬t) of the particle at the end of the time interval ⌬t is approximately equal to the speed v(t) at the beginning of the time interval plus the
magnitude of the acceleration during the interval multiplied by ⌬t:
v(t ϩ ⌬t) Ϸ v(t) ϩ a(t)⌬t

(6.10)

Because the acceleration is a function of time, this estimate of v(t ϩ ⌬t) is accurate
only if the time interval ⌬t is short enough that the change in acceleration during
it is very small (as is discussed later). Of course, Equation 6.10 is exact if the acceleration is constant.


6.5

Numerical Modeling in Particle Dynamics

171

The position x(t ϩ ⌬t) of the particle at the end of the interval ⌬t can be
found in the same manner:

v(t) Ϸ

⌬x
x(t ϩ ⌬t) Ϫ x(t)
ϭ
⌬t
⌬t

x(t ϩ ⌬t) Ϸ x(t) ϩ v(t)⌬t
1
2

(6.11)

a(⌬t)2

You may be tempted to add the term
to this result to make it look like
the familiar kinematics equation, but this term is not included in the Euler
method because ⌬t is assumed to be so small that ⌬t 2 is nearly zero.
If the acceleration at any instant t is known, the particle’s velocity and position
at a time t ϩ ⌬t can be calculated from Equations 6.10 and 6.11. The calculation
then proceeds in a series of finite steps to determine the velocity and position at
any later time. The acceleration is determined from the net force acting on the
particle, and this force may depend on position, velocity, or time:
a(x, v, t) ϭ

⌺ F(x, v, t)

(6.12)

m

It is convenient to set up the numerical solution to this kind of problem by
numbering the steps and entering the calculations in a table, a procedure that is illustrated in Table 6.3.
The equations in the table can be entered into a spreadsheet and the calculations performed row by row to determine the velocity, position, and acceleration
as functions of time. The calculations can also be carried out by using a program
written in either BASIC, Cϩϩ, or FORTRAN or by using commercially available
mathematics packages for personal computers. Many small increments can be
taken, and accurate results can usually be obtained with the help of a computer.
Graphs of velocity versus time or position versus time can be displayed to help you
visualize the motion.
One advantage of the Euler method is that the dynamics is not obscured — the
fundamental relationships between acceleration and force, velocity and acceleration, and position and velocity are clearly evident. Indeed, these relationships
form the heart of the calculations. There is no need to use advanced mathematics,
and the basic physics governs the dynamics.
The Euler method is completely reliable for infinitesimally small time increments, but for practical reasons a finite increment size must be chosen. For the finite difference approximation of Equation 6.10 to be valid, the time increment
must be small enough that the acceleration can be approximated as being constant during the increment. We can determine an appropriate size for the time in-

TABLE 6.3 The Euler Method for Solving Dynamics Problems
Step
0
1
2
3
n

Time

Position

Velocity

Acceleration

t0
t 1 ϭ t 0 ϩ ⌬t
t 2 ϭ t 1 ϩ ⌬t
t 3 ϭ t 2 ϩ ⌬t
Ӈ
tn

x0
x 1 ϭ x 0 ϩ v0 ⌬t
x 2 ϭ x 1 ϩ v 1 ⌬t
x 3 ϭ x 2 ϩ v 2 ⌬t
Ӈ
xn

v0
v 1 ϭ v0 ϩ a 0 ⌬t
v 2 ϭ v 1 ϩ a 1 ⌬t
v 3 ϭ v 2 ϩ a 2 ⌬t
Ӈ
vn

a 0 ϭ F(x 0 , v0 , t 0)/m
a 1 ϭ F(x 1 , v 1 , t 1)/m
a 2 ϭ F(x 2 , v 2 , t 2)/m
a 3 ϭ F(x 3 , v 3 , t 3)/m

Ӈ
an

See the spreadsheet file “Baseball
with Drag” on the Student Web
site (address below) for an
example of how this technique can
be applied to find the initial speed
of the baseball described in
Example 6.14. We cannot use our
regular approach because our
kinematics equations assume
constant acceleration. Euler’s
method provides a way to
circumvent this difficulty.

A detailed solution to Problem 41
involving iterative integration
appears in the Student Solutions
Manual and Study Guide and is
posted on the Web at http:/
www.saunderscollege.com/physics


172

CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws

crement by examining the particular problem being investigated. The criterion for
the size of the time increment may need to be changed during the course of the
motion. In practice, however, we usually choose a time increment appropriate to
the initial conditions and use the same value throughout the calculations.
The size of the time increment influences the accuracy of the result, but unfortunately it is not easy to determine the accuracy of an Euler-method solution
without a knowledge of the correct analytical solution. One method of determining the accuracy of the numerical solution is to repeat the calculations with a
smaller time increment and compare results. If the two calculations agree to a certain number of significant figures, you can assume that the results are correct to
that precision.

SUMMARY
Newton’s second law applied to a particle moving in uniform circular motion states
that the net force causing the particle to undergo a centripetal acceleration is

⌺ Fr ϭ mar ϭ

mv2
r

(6.1)

You should be able to use this formula in situations where the force providing the
centripetal acceleration could be the force of gravity, a force of friction, a force of
string tension, or a normal force.
A particle moving in nonuniform circular motion has both a centripetal component of acceleration and a nonzero tangential component of acceleration. In
the case of a particle rotating in a vertical circle, the force of gravity provides the
tangential component of acceleration and part or all of the centripetal component
of acceleration. Be sure you understand the directions and magnitudes of the velocity and acceleration vectors for nonuniform circular motion.
An observer in a noninertial (accelerating) frame of reference must introduce
fictitious forces when applying Newton’s second law in that frame. If these fictitious forces are properly defined, the description of motion in the noninertial
frame is equivalent to that made by an observer in an inertial frame. However, the

observers in the two frames do not agree on the causes of the motion. You should
be able to distinguish between inertial and noninertial frames and identify the fictitious forces acting in a noninertial frame.
A body moving through a liquid or gas experiences a resistive force that is
speed-dependent. This resistive force, which opposes the motion, generally increases with speed. The magnitude of the resistive force depends on the shape of
the body and on the properties of the medium through which the body is moving.
In the limiting case for a falling body, when the magnitude of the resistive force
equals the body’s weight, the body reaches its terminal speed. You should be able
to apply Newton’s laws to analyze the motion of objects moving under the influence of resistive forces. You may need to apply Euler’s method if the force depends on velocity, as it does for air drag.

QUESTIONS
1. Because the Earth rotates about its axis and revolves
around the Sun, it is a noninertial frame of reference. Assuming the Earth is a uniform sphere, why would the ap-

parent weight of an object be greater at the poles than at
the equator?
2. Explain why the Earth bulges at the equator.


Problems
3. Why is it that an astronaut in a space capsule orbiting the
Earth experiences a feeling of weightlessness?
4. Why does mud fly off a rapidly turning automobile tire?
5. Imagine that you attach a heavy object to one end of a
spring and then whirl the spring and object in a horizontal circle (by holding the free end of the spring). Does
the spring stretch? If so, why? Discuss this in terms of the
force causing the circular motion.
6. It has been suggested that rotating cylinders about 10 mi
in length and 5 mi in diameter be placed in space and
used as colonies. The purpose of the rotation is to simulate gravity for the inhabitants. Explain this concept for
producing an effective gravity.

7. Why does a pilot tend to black out when pulling out of a
steep dive?

173

8. Describe a situation in which a car driver can have
a centripetal acceleration but no tangential acceleration.
9. Describe the path of a moving object if its acceleration is
constant in magnitude at all times and (a) perpendicular
to the velocity; (b) parallel to the velocity.
10. Analyze the motion of a rock falling through water in
terms of its speed and acceleration as it falls. Assume that
the resistive force acting on the rock increases as the
speed increases.
11. Consider a small raindrop and a large raindrop falling
through the atmosphere. Compare their terminal speeds.
What are their accelerations when they reach terminal
speed?

PROBLEMS
1, 2, 3 = straightforward, intermediate, challenging
= full solution available in the Student Solutions Manual and Study Guide
WEB = solution posted at />= Computer useful in solving problem
= Interactive Physics
= paired numerical/symbolic problems

Newton’s Second Law
Applied to Uniform Circular Motion

Section 6.1

1. A toy car moving at constant speed completes one lap
around a circular track (a distance of 200 m) in 25.0 s.
(a) What is its average speed? (b) If the mass of the car
is 1.50 kg, what is the magnitude of the force that keeps
it in a circle?
2. A 55.0-kg ice skater is moving at 4.00 m/s when she
grabs the loose end of a rope, the opposite end of
which is tied to a pole. She then moves in a circle of radius 0.800 m around the pole. (a) Determine the force
exerted by the rope on her arms. (b) Compare this
force with her weight.
3. A light string can support a stationary hanging load of
25.0 kg before breaking. A 3.00-kg mass attached to the
string rotates on a horizontal, frictionless table in a circle of radius 0.800 m. What range of speeds can the
mass have before the string breaks?
4. In the Bohr model of the hydrogen atom, the speed of
the electron is approximately 2.20 ϫ 106 m/s. Find
(a) the force acting on the electron as it revolves in a
circular orbit of radius 0.530 ϫ 10Ϫ10 m and (b) the
centripetal acceleration of the electron.
5. In a cyclotron (one type of particle accelerator), a
deuteron (of atomic mass 2.00 u) reaches a final speed
of 10.0% of the speed of light while moving in a circular
path of radius 0.480 m. The deuteron is maintained in
the circular path by a magnetic force. What magnitude
of force is required?
6. A satellite of mass 300 kg is in a circular orbit around
the Earth at an altitude equal to the Earth’s mean radius (see Example 6.6). Find (a) the satellite’s orbital

7.

8.

9.

10.

11.

speed, (b) the period of its revolution, and (c) the gravitational force acting on it.
Whenever two Apollo astronauts were on the surface of
the Moon, a third astronaut orbited the Moon. Assume
the orbit to be circular and 100 km above the surface of
the Moon. If the mass of the Moon is 7.40 ϫ 1022 kg and
its radius is 1.70 ϫ 106 m, determine (a) the orbiting astronaut’s acceleration, (b) his orbital speed, and (c) the
period of the orbit.
The speed of the tip of the minute hand on a town
clock is 1.75 ϫ 10Ϫ3 m/s. (a) What is the speed of the
tip of the second hand of the same length? (b) What is
the centripetal acceleration of the tip of the second
hand?
A coin placed 30.0 cm from the center of a rotating,
horizontal turntable slips when its speed is 50.0 cm/s.
(a) What provides the force in the radial direction
when the coin is stationary relative to the turntable?
(b) What is the coefficient of static friction between
coin and turntable?
The cornering performance of an automobile is evaluated on a skid pad, where the maximum speed that a
car can maintain around a circular path on a dry, flat
surface is measured. The centripetal acceleration, also

called the lateral acceleration, is then calculated as a
multiple of the free-fall acceleration g. The main factors
affecting the performance are the tire characteristics
and the suspension system of the car. A Dodge Viper
GTS can negotiate a skid pad of radius 61.0 m at
86.5 km/h. Calculate its maximum lateral acceleration.
A crate of eggs is located in the middle of the flatbed of
a pickup truck as the truck negotiates an unbanked


174

CHAPTER 6

Circular Motion and Other Applications of Newton’s Laws

curve in the road. The curve may be regarded as an arc
of a circle of radius 35.0 m. If the coefficient of static
friction between crate and truck is 0.600, how fast can
WEB
the truck be moving without the crate sliding?
12. A car initially traveling eastward turns north by traveling
in a circular path at uniform speed as in Figure P6.12.
The length of the arc ABC is 235 m, and the car completes the turn in 36.0 s. (a) What is the acceleration
when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors i and j.
Determine (b) the car ’s average speed and (c) its average acceleration during the 36.0-s interval.

hump? (b) What must be the speed of the car over the
hump if she is to experience weightlessness? (That is, if
her apparent weight is zero.)

15. Tarzan (m ϭ 85.0 kg) tries to cross a river by swinging
from a vine. The vine is 10.0 m long, and his speed at
the bottom of the swing (as he just clears the water) is
8.00 m/s. Tarzan doesn’t know that the vine has a
breaking strength of 1 000 N. Does he make it safely
across the river?
16. A hawk flies in a horizontal arc of radius 12.0 m at a
constant speed of 4.00 m/s. (a) Find its centripetal acceleration. (b) It continues to fly along the same horizontal arc but steadily increases its speed at the rate of
1.20 m/s2. Find the acceleration (magnitude and direction) under these conditions.

y
O

35.0°

C

17. A 40.0-kg child sits in a swing supported by two chains,
each 3.00 m long. If the tension in each chain at the
lowest point is 350 N, find (a) the child’s speed at the
lowest point and (b) the force exerted by the seat on
the child at the lowest point. (Neglect the mass of the
seat.)
18. A child of mass m sits in a swing supported by two
chains, each of length R. If the tension in each chain at
the lowest point is T, find (a) the child’s speed at the
lowest point and (b) the force exerted by the seat on
the child at the lowest point. (Neglect the mass of the
seat.)

x

B
A

Figure P6.12
13. Consider a conical pendulum with an 80.0-kg bob on a
10.0-m wire making an angle of ␪ ϭ 5.00° with the vertical (Fig. P6.13). Determine (a) the horizontal and vertical components of the force exerted by the wire on the
pendulum and (b) the radial acceleration of the bob.

θ

WEB

19. A pail of water is rotated in a vertical circle of radius
1.00 m. What must be the minimum speed of the pail at
the top of the circle if no water is to spill out?
20. A 0.400-kg object is swung in a vertical circular path on
a string 0.500 m long. If its speed is 4.00 m/s at the top
of the circle, what is the tension in the string there?
21. A roller-coaster car has a mass of 500 kg when fully
loaded with passengers (Fig. P6.21). (a) If the car has a
speed of 20.0 m/s at point A, what is the force exerted
by the track on the car at this point? (b) What is the
maximum speed the car can have at B and still remain
on the track?

B

15.0 m

Figure P6.13
Section 6.2

10.0 m
A

Nonuniform Circular Motion

14. A car traveling on a straight road at 9.00 m/s goes over
a hump in the road. The hump may be regarded as an
arc of a circle of radius 11.0 m. (a) What is the apparent
weight of a 600-N woman in the car as she rides over the

Figure P6.21


175

Problems
22. A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some of the
latest design technology and some basic physics. Each
vertical loop, instead of being circular, is shaped like a
teardrop (Fig. P6.22). The cars ride on the inside of the
loop at the top, and the speeds are high enough to ensure that the cars remain on the track. The biggest loop
is 40.0 m high, with a maximum speed of 31.0 m/s
(nearly 70 mi/h) at the bottom. Suppose the speed at
the top is 13.0 m/s and the corresponding centripetal
acceleration is 2g. (a) What is the radius of the arc of
the teardrop at the top? (b) If the total mass of the cars

plus people is M, what force does the rail exert on this
total mass at the top? (c) Suppose the roller coaster had
a loop of radius 20.0 m. If the cars have the same speed,
13.0 m/s at the top, what is the centripetal acceleration
at the top? Comment on the normal force at the top in
this situation.

24. A 5.00-kg mass attached to a spring scale rests on a frictionless, horizontal surface as in Figure P6.24. The
spring scale, attached to the front end of a boxcar, reads
18.0 N when the car is in motion. (a) If the spring scale
reads zero when the car is at rest, determine the acceleration of the car. (b) What will the spring scale read if
the car moves with constant velocity? (c) Describe the
forces acting on the mass as observed by someone in
the car and by someone at rest outside the car.

5.00 kg

Figure P6.24

Figure P6.22

(Frank Cezus/FPG International)

(Optional)

Section 6.3

Motion in Accelerated Frames

23. A merry-go-round makes one complete revolution in

12.0 s. If a 45.0-kg child sits on the horizontal floor of
the merry-go-round 3.00 m from the center, find (a) the
child’s acceleration and (b) the horizontal force of friction that acts on the child. (c) What minimum coefficient of static friction is necessary to keep the child
from slipping?

25. A 0.500-kg object is suspended from the ceiling of an
accelerating boxcar as was seen in Figure 6.13. If a ϭ
3.00 m/s2, find (a) the angle that the string makes with
the vertical and (b) the tension in the string.
26. The Earth rotates about its axis with a period of 24.0 h.
Imagine that the rotational speed can be increased. If
an object at the equator is to have zero apparent weight,
(a) what must the new period be? (b) By what factor
would the speed of the object be increased when the
planet is rotating at the higher speed? (Hint: See Problem 53 and note that the apparent weight of the object
becomes zero when the normal force exerted on it is
zero. Also, the distance traveled during one period is
2␲R, where R is the Earth’s radius.)
27. A person stands on a scale in an elevator. As the elevator
starts, the scale has a constant reading of 591 N. As the
elevator later stops, the scale reading is 391 N. Assume
the magnitude of the acceleration is the same during
starting and stopping, and determine (a) the weight of
the person, (b) the person’s mass, and (c) the acceleration of the elevator.
28. A child on vacation wakes up. She is lying on her back.
The tension in the muscles on both sides of her neck is
55.0 N as she raises her head to look past her toes and
out the motel window. Finally, it is not raining! Ten minutes later she is screaming and sliding feet first down a
water slide at a constant speed of 5.70 m/s, riding high
on the outside wall of a horizontal curve of radius 2.40 m

(Fig. P6.28). She raises her head to look forward past
her toes; find the tension in the muscles on both sides
of her neck.